When iron interacted with a solution of copper (II) sulfate, 25.4 g of copper were released.

When iron interacted with a solution of copper (II) sulfate, 25.4 g of copper were released. How many grams of technical iron, containing 3% impurities, was involved in the reaction?

1. More active metals displace less active metals from solutions of their salts, we write down the reaction equation:

Fe + CuSO4 = Cu + FeSO4;

2.Calculate the chemical amount of released copper:

n (Cu) = m (Cu): M (Cu) = 25.4: 64 = 0.3969 mol;

3. Determine the chemical amount of iron:

n (Fe) = n (Cu) = 0.3969 mol;

4.calculate the mass of iron:

m (Fe) = n (Fe) * M (Fe) = 0.3969 * 56 = 22.23 g;

5.Calculate the mass of a technical sample of iron:

w (Fe) = 1 – w (impurities) = 1 – 0.03 = 0.97;

mtech (Fe) = m (Fe): w (Fe) = 22.23: 0.97 = 22.92 g.

Answer: 22.92 g.